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Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is 8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

Answers

Answer 1

Answer:

Answer:

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Explanation:

Inductance of a solenoid is

$L=N\frac\phi I$

$=N(B.A)/(I)$

$=N(\mu_0NI)/(l.I)A$

$=(\mu_0N^2A)/(l)$

$=(\mu_0N^2)/(l).\pi(\frac d2)^2$

$=\mu_0\pi(N^2d^2)/(4l)$

N= number of turns

$l$ = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

$\phi$ = magnetic flux

$I$ = Current

Given that, Solenoid A has total number of turns N, length L and diameter D

The inductance of solenoid A is

$=\mu_0\pi(N^2D^2)/(4L)$

Solenoid B has total number of turns 2N, length 2L and diameter 2D

The inductance of solenoid B is

$=\mu_0\pi((2N)^2(2D)^2)/(4.2L)$

$=\mu_0\pi(16 N^2D^2)/(4.2L)$

Therefore,

$\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=(\mu_0\pi(N^2D^2)/(4L))/(\mu_0\pi(16 N^2D^2)/(4.2L))$

$\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18$

$\Rightarrow {\textrm{Inductance of A}}=\frac18* {\textrm{Inductance of B}}$

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Answer 2

Answer:

Hi there!

We can begin by calculating the inductance of a solenoid.

Recall:
$L = (\Phi _B)/(i)$

L = Inductance (H)
φ = Magnetic Flux (Wb)

i = Current (A)

We can solve for the inductance of a solenoid. We know that its magnetic field is equivalent to:
$B = \mu _0 (N)/(L)i$

And that the magnetic flux is equivalent to:
$\Phi _B = \int B \cdot dA = B \cdot A$

Thus, the magnetic flux is equivalent to:
$\Phi _B = \mu _0 (N)/(L)iA$

The area for the solenoid is the # of loops multiplied by the cross-section area, so:
$A_(total)= N * A$

$\Phi _B = \mu _0 (N^2)/(L)iA$

Using this equation, we can find how it would change if the given parameters are altered:
$\Phi_B ' = \mu_0 ((2N)^2)/(2L) i * 4A$

**The area will quadruple since a circle's area is 2-D, and you are doubling its diameter.

$\Phi'_B = (4)/(2) * 4(\mu_0 (N)/(L)iA) = 8\mu_0 (N)/(L)iA$

Thus, Solenoid B is 8 times as large as Solenoid A.

Solenoid A is 1/8 of the inductance of solenoid B.

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The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

Answers

Answer:

Charge, $Q=1.56* 10^(-8)\ C$

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

$E=(kQ)/(r^2)$

Q is the charge on an object

$Q=(Er^2)/(k)$

$Q=(180000\ N/C* (0.028\ m)^2)/(9* 10^9\ Nm^2/C^2)$

$Q=1.56* 10^(-8)\ C$

So, the charge on the object is $1.56* 10^(-8)\ C$ . Hence, this is the required solution.

You are measuring the volume of a chemical beaker how would u take the measurment? a.lift the beaker to eye level
b.you look down at the liquid above
c.use a ruler to measure
d.you squat to be eye level to the beaker

Answers

I think its d because lifting it would make the chemical swish around and that will make it so you cant get the right measurement. hope this helps :)

What happens at night- describing air circulation

Answers

Answer:

The environment is warmed by the light throughout the day, such that the temperature increases. The weather is decreasing and the temperature decreases in the night as the sun falls. There was a misunderstanding. Thanks to the density, the atmosphere becomes densest on the earth. The air becomes colder and colder when you move up.

Explanation:

Answer is above

Hope this helps.

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1?
B. What is the change in the total momentum of the pair?
C. What is the magnitude of the change in the momentum Δp2, of mass M2?

Answers

Answer:

a). ΔP1=-2.4 $x10^(3) (D*m)/(s)$

b). Pp=0 F=0

c). ΔP2=2.4 $x10^(3) (D*m)/(s)$

Explanation:

Initial momentum

$P_(1)=m_(1)*v_(i1)$

Final momentum

$P_(1f)=m_(1)*v_(f1)=-m_(1)*v_(i1)$

The change of momentum m1 is:

a).

ΔP1= $P_(1f)-P_(1)$

ΔP1= $-m_(1)*v_(i1)-m_(1)*v_(i1)$

ΔP1= $-2*m_(1)*v_(i1)$

ΔP1= $-2*6 D*200(m)/(s)$

ΔP1= $-2.4x10^(3)(D*m)/(s)$

b).

The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero

P=0

Fx=0

c).

ΔP1+ΔP2=0

ΔP2=-ΔP1

ΔP2=- $-2.4x10^(3)(D*m)/(s)$

ΔP2= $2.4x10^(3)(D*m)/(s)$

Final answer:

The magnitude of the change in momentum of mass M1 is 2400 Daltons*m/s. The change in the total momentum of the pair is 2000 Daltons*m/s. The magnitude of the change in momentum of mass M2 is -400 Daltons*m/s.

Explanation:

A. To find the magnitude of the change in momentum of mass M1, we use the formula Δp1 = m1 * Δv1, where m1 is the mass of M1 and Δv1 is the change in velocity of M1. Since M1 simply changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, Δp1 = m1 * (2v1) = 6 * (2 * 200) = 2400 Daltons*m/s.

B. The change in the total momentum of the pair is equal to the sum of the changes in momentum of M1 and M2. Since M2 also changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, the change in the total momentum is Δp1 + Δp2 = 2400 Daltons*m/s + (-400 Daltons*m/s) = 2000 Daltons*m/s.

C. To find the magnitude of the change in momentum of mass M2, we use the same formula as in part A, but with the values for M2. Δp2 = m2 * Δv2 = 1 * (2 * (-200)) = -400 Daltons*m/s.

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A parallel-plate capacitor is charged and then disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?

Answers

Answer:

U/U₀ = 2

(factor of 2 i.e U = 2U₀)

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected

Explanation:

Energy stored in a capacitor can be expressed as;

U = 0.5CV^2 = Q^2/2C

And

C = ε₀ A/d

Where

C = capacitance

V = potential difference

Q = charge

A = Area of plates

d = distance between plates

U = Q^2/2C = dQ^2/2ε₀ A

The initial energy of the capacitor at d = d₀ is

U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1

When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.

The final energy stored in the capacitor at d = 2d₀ is

U = 2d₀Q^2/2ε₀ A ...2

The factor U/U₀ can be derived by substituting equation 1 and 2

U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )

Simplifying we have;

U/U₀ = 2

U = 2U₀

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.

Anatomy of a Wave worksheet can someone help me out with the answers????

Answers

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